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Maslowich
1 year ago
7

During the day, susan notices that the wind is blowing onshore at the beach. What is this called? land breeze land breeze sea br

eeze sea breeze monsoon monsoon global wind
Physics
1 answer:
In-s [12.5K]1 year ago
8 0

The correct option is (b) Sea breeze

During the day, susan notices that the wind is blowing onshore at the beach called sea breeze.

What is sea breeze?

  • Any wind that flows from a big body of water onto or onto a landmass is called a sea breeze or an onshore breeze.
  • Sea breezes form as a result of changes in air pressure brought on by the different heat capacities of water and dry land. Sea breezes are therefore more confined than prevailing winds.
  • A sea wind is frequently seen along coasts after sunrise because land warms up far more quickly than water does when exposed to solar radiation.
  • The sea wind front is significant because it can serve as a catalyst for afternoon thunderstorms and provide welcome cooling along the coast.

Learn more about the sea breeze with the help of the given link:

brainly.com/question/13015619

#SPJ4

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Can anyone check if my answer is correct ?
ohaa [14]

I believe your answer is correct, because 8.7*10^-7 is equal to 0.00000085347.

Hope you do well!

4 0
2 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
2 years ago
How does superposition provide evidence for the evolution of earth?
77julia77 [94]

Answer:

Explanation: This Law of Superposition is fundamental to the interpretation of Earth history, because at any one location it indicates the relative ages of rock layers and the fossils in them.

7 0
3 years ago
A cannon fired horizontally at 20 m/s from the top of a cliff lands 80m away. how tall is the cliff
Scorpion4ik [409]

Answer:

The height of the cliff is, h = 78.4 m

Explanation:

Given,

The horizontal velocity of the projectile, Vx = 20 m/s

The range of the projectile, s = 80 m

The projectile projected from a height is given by the formula

                            <em> S = Vx [Vy + √(Vy² + 2gh)] / g </em>

Therefore,  

                            h = S²g/2Vx²

Substituting the values

                             h = 80² x 9.8/ (2 x 20²)

                                = 78.4 m

Hence, the height of the cliff is, h = 78.4 m

8 0
3 years ago
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete
disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) F_{net} \text {on wire }3=0

\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm

position of wire = 50 - 1.2

= 48.8cm

b)  F_{net} \text {on wire }1=0

\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A

Direction ⇒ downward

5 0
3 years ago
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